Mixing
What is the limit of $lim_{nto infty} (1 - frac{1}{4})(1 - frac{1}{9})(1 - frac{1}{16}) cdots (1 -...
What is the evaluation of the following infinite series?
$$lim_{nto infty} left(1 - frac{1}{4}right)left(1 - frac{1}{9}right)left(1 - frac{1}{16}right) cdots left(1- frac{1}{(n+1)^2}right)$$
I've tried to simplify each expression which left me with:
$$lim_{nto infty} frac{3times8times15times24timescdotstimes((n+1)^2-1)}{4times9times16times25timescdotstimes(n+1)^2}$$
Is this a good way to approach this problem?
sequences-and-series limits
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
add a comment |
What is the evaluation of the following infinite series?
$$lim_{nto infty} left(1 - frac{1}{4}right)left(1 - frac{1}{9}right)left(1 - frac{1}{16}right) cdots left(1- frac{1}{(n+1)^2}right)$$
I've tried to simplify each expression which left me with:
$$lim_{nto infty} frac{3times8times15times24timescdotstimes((n+1)^2-1)}{4times9times16times25timescdotstimes(n+1)^2}$$
Is this a good way to approach this problem?
sequences-and-series limits
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
3
I'd try $(1-frac1{n^2})=(1-frac1 n)(1+frac1 n)$ then try just partial product $prod_{n=2}^m (1-frac 1n)$ then partial product $prod_{n=2}^m(1+frac 1n)$ then combine the partial products
– kingW3
Nov 20 '18 at 16:49
2
Your product omits $left(1-frac1{16}right)$; is this an error, or is it meant to be e.g. over only primes?
– Steven Stadnicki
Nov 20 '18 at 16:49
@StevenStadnicki You're correct. Fixed
– Stefan Ivanovski
Nov 20 '18 at 16:51
add a comment |
What is the evaluation of the following infinite series?
$$lim_{nto infty} left(1 - frac{1}{4}right)left(1 - frac{1}{9}right)left(1 - frac{1}{16}right) cdots left(1- frac{1}{(n+1)^2}right)$$
I've tried to simplify each expression which left me with:
$$lim_{nto infty} frac{3times8times15times24timescdotstimes((n+1)^2-1)}{4times9times16times25timescdotstimes(n+1)^2}$$
Is this a good way to approach this problem?
sequences-and-series limits
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
What is the evaluation of the following infinite series?
$$lim_{nto infty} left(1 - frac{1}{4}right)left(1 - frac{1}{9}right)left(1 - frac{1}{16}right) cdots left(1- frac{1}{(n+1)^2}right)$$
I've tried to simplify each expression which left me with:
$$lim_{nto infty} frac{3times8times15times24timescdotstimes((n+1)^2-1)}{4times9times16times25timescdotstimes(n+1)^2}$$
Is this a good way to approach this problem?
sequences-and-series limits
sequences-and-series limits
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
edited Nov 20 '18 at 17:06
Lorenzo B.
1,8322520
1,8322520
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
asked Nov 20 '18 at 16:45
Stefan Ivanovski
546
546
3
I'd try $(1-frac1{n^2})=(1-frac1 n)(1+frac1 n)$ then try just partial product $prod_{n=2}^m (1-frac 1n)$ then partial product $prod_{n=2}^m(1+frac 1n)$ then combine the partial products
– kingW3
Nov 20 '18 at 16:49
2
Your product omits $left(1-frac1{16}right)$; is this an error, or is it meant to be e.g. over only primes?
– Steven Stadnicki
Nov 20 '18 at 16:49
@StevenStadnicki You're correct. Fixed
– Stefan Ivanovski
Nov 20 '18 at 16:51
add a comment |
3
I'd try $(1-frac1{n^2})=(1-frac1 n)(1+frac1 n)$ then try just partial product $prod_{n=2}^m (1-frac 1n)$ then partial product $prod_{n=2}^m(1+frac 1n)$ then combine the partial products
– kingW3
Nov 20 '18 at 16:49
2
Your product omits $left(1-frac1{16}right)$; is this an error, or is it meant to be e.g. over only primes?
– Steven Stadnicki
Nov 20 '18 at 16:49
@StevenStadnicki You're correct. Fixed
– Stefan Ivanovski
Nov 20 '18 at 16:51
3
3
I'd try $(1-frac1{n^2})=(1-frac1 n)(1+frac1 n)$ then try just partial product $prod_{n=2}^m (1-frac 1n)$ then partial product $prod_{n=2}^m(1+frac 1n)$ then combine the partial products
– kingW3
Nov 20 '18 at 16:49
I'd try $(1-frac1{n^2})=(1-frac1 n)(1+frac1 n)$ then try just partial product $prod_{n=2}^m (1-frac 1n)$ then partial product $prod_{n=2}^m(1+frac 1n)$ then combine the partial products
– kingW3
Nov 20 '18 at 16:49
2
2
Your product omits $left(1-frac1{16}right)$; is this an error, or is it meant to be e.g. over only primes?
– Steven Stadnicki
Nov 20 '18 at 16:49
Your product omits $left(1-frac1{16}right)$; is this an error, or is it meant to be e.g. over only primes?
– Steven Stadnicki
Nov 20 '18 at 16:49
@StevenStadnicki You're correct. Fixed
– Stefan Ivanovski
Nov 20 '18 at 16:51
@StevenStadnicki You're correct. Fixed
– Stefan Ivanovski
Nov 20 '18 at 16:51
add a comment |
6 Answers
6
active
oldest
votes
You want
begin{align*}prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right)
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)left(1+frac{1}{k+1}right) \
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)prod_{k=1}^{n} left(1+frac{1}{k+1}right) tag{1}end{align*}
Now,
$$
prod_{k=1}^{n} left(1-frac{1}{k+1}right) = prod_{k=1}^{n} frac{k}{k+1} = frac{1}{n+1}tag{2}
$$
(as things cancel out/telescope), and similarly
$$
prod_{k=1}^{n} left(1+frac{1}{k+1}right) = prod_{k=1}^{n} frac{k+2}{k+1} = frac{n+2}{2} tag{3}
$$
and so, combining (2) and (3) into (1),
$$
prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right) = frac{1}{2}cdotfrac{n+2}{n+1}
$$
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
add a comment |
$$prod_{k=1}^ndfrac{f(k)}{f(k+1)}=dfrac{f(1)}{f(n+1)}$$
Here $f(m)=dfrac m{m+1}$
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
|
show 2 more comments
$frac{1times 3}{2times 2}cdotfrac{2times 4}{3times 3}cdotfrac{3times 5}{4times 4}cdotfrac{4times 6}{5times 5}cdots$
Do you see it?
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
add a comment |
hint
Take logarithm and use
$$
sum_{k=2}^{n+1}lnleft(1-frac{1}{k^2}right)
= sum_{k=2}^{n+1}left(lnleft(frac{k-1}{k}right)+lnleft(frac{k+1}{k}right)right)
=lnleft(frac{1}{n+1}right)+lnleft(frac{n+2}{2}right).
$$
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
add a comment |
Since
$$frac{sin(pi x)}{pi x}=prod_{ngeq 1}left(1-frac{x^2}{n^2}right) $$
we simply have
$$ prod_{ngeq 2}left(1-frac{1}{n^2}right)=lim_{xto 1}frac{sin(pi x)}{pi x(1-x^2)}=lim_{zto 0}frac{-sin(pi z)}{-pi z(z+1)(z+2)}=color{red}{frac{1}{2}}. $$
This technique is interesting since it allows a quick evaluation of $prod_{ngeq 1}left(1-frac{1}{2n^2}right)$ and similar products, which are not telescopic.
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{prod_{k = 1}^{n}bracks{1 - {1 over pars{k + 1}^{2}}}} =
overbrace{{prod_{k = 1}^{n}k over prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {1 over n + 1}}}
overbrace{{prod_{k = 1}^{n}pars{k + 2} over
prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {n + 2 over 2}}} \[5mm] = &
{1 over 2},{1 + 2/n over 1 + 1/n}
,,,stackrel{mrm{as} n to infty}
{Large to},,,bbx{1 over 2}
end{align}
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
You want
begin{align*}prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right)
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)left(1+frac{1}{k+1}right) \
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)prod_{k=1}^{n} left(1+frac{1}{k+1}right) tag{1}end{align*}
Now,
$$
prod_{k=1}^{n} left(1-frac{1}{k+1}right) = prod_{k=1}^{n} frac{k}{k+1} = frac{1}{n+1}tag{2}
$$
(as things cancel out/telescope), and similarly
$$
prod_{k=1}^{n} left(1+frac{1}{k+1}right) = prod_{k=1}^{n} frac{k+2}{k+1} = frac{n+2}{2} tag{3}
$$
and so, combining (2) and (3) into (1),
$$
prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right) = frac{1}{2}cdotfrac{n+2}{n+1}
$$
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
add a comment |
You want
begin{align*}prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right)
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)left(1+frac{1}{k+1}right) \
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)prod_{k=1}^{n} left(1+frac{1}{k+1}right) tag{1}end{align*}
Now,
$$
prod_{k=1}^{n} left(1-frac{1}{k+1}right) = prod_{k=1}^{n} frac{k}{k+1} = frac{1}{n+1}tag{2}
$$
(as things cancel out/telescope), and similarly
$$
prod_{k=1}^{n} left(1+frac{1}{k+1}right) = prod_{k=1}^{n} frac{k+2}{k+1} = frac{n+2}{2} tag{3}
$$
and so, combining (2) and (3) into (1),
$$
prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right) = frac{1}{2}cdotfrac{n+2}{n+1}
$$
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
add a comment |
You want
begin{align*}prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right)
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)left(1+frac{1}{k+1}right) \
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)prod_{k=1}^{n} left(1+frac{1}{k+1}right) tag{1}end{align*}
Now,
$$
prod_{k=1}^{n} left(1-frac{1}{k+1}right) = prod_{k=1}^{n} frac{k}{k+1} = frac{1}{n+1}tag{2}
$$
(as things cancel out/telescope), and similarly
$$
prod_{k=1}^{n} left(1+frac{1}{k+1}right) = prod_{k=1}^{n} frac{k+2}{k+1} = frac{n+2}{2} tag{3}
$$
and so, combining (2) and (3) into (1),
$$
prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right) = frac{1}{2}cdotfrac{n+2}{n+1}
$$
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
You want
begin{align*}prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right)
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)left(1+frac{1}{k+1}right) \
& = prod_{k=1}^{n} left(1-frac{1}{k+1}right)prod_{k=1}^{n} left(1+frac{1}{k+1}right) tag{1}end{align*}
Now,
$$
prod_{k=1}^{n} left(1-frac{1}{k+1}right) = prod_{k=1}^{n} frac{k}{k+1} = frac{1}{n+1}tag{2}
$$
(as things cancel out/telescope), and similarly
$$
prod_{k=1}^{n} left(1+frac{1}{k+1}right) = prod_{k=1}^{n} frac{k+2}{k+1} = frac{n+2}{2} tag{3}
$$
and so, combining (2) and (3) into (1),
$$
prod_{k=1}^{n} left(1-frac{1}{(k+1)^2}right) = frac{1}{2}cdotfrac{n+2}{n+1}
$$
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
edited Nov 20 '18 at 17:16
Viktor Glombik
582425
582425
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
answered Nov 20 '18 at 16:57
Clement C.
49.7k33886
49.7k33886
add a comment |
add a comment |
$$prod_{k=1}^ndfrac{f(k)}{f(k+1)}=dfrac{f(1)}{f(n+1)}$$
Here $f(m)=dfrac m{m+1}$
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
|
show 2 more comments
$$prod_{k=1}^ndfrac{f(k)}{f(k+1)}=dfrac{f(1)}{f(n+1)}$$
Here $f(m)=dfrac m{m+1}$
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
|
show 2 more comments
$$prod_{k=1}^ndfrac{f(k)}{f(k+1)}=dfrac{f(1)}{f(n+1)}$$
Here $f(m)=dfrac m{m+1}$
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
$$prod_{k=1}^ndfrac{f(k)}{f(k+1)}=dfrac{f(1)}{f(n+1)}$$
Here $f(m)=dfrac m{m+1}$
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
answered Nov 20 '18 at 17:10
lab bhattacharjee
223k15156274
223k15156274
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
|
show 2 more comments
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
Your answers are very short but very much understandable @lab bhattacharjee. Please suggest me some books of Maths so that I can also become a Math genius. Currently I have nearly completed the Vinay Kumar tmh books and am in class 12.
– Akash Roy
Nov 20 '18 at 17:14
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
google.co.in/…
– lab bhattacharjee
Nov 20 '18 at 18:17
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@Akash, archive.org/details/higheralgebra032813mbp
– lab bhattacharjee
Nov 20 '18 at 18:18
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
@AkashRoy, See google.co.in/…
– lab bhattacharjee
Nov 21 '18 at 8:56
|
show 2 more comments
$frac{1times 3}{2times 2}cdotfrac{2times 4}{3times 3}cdotfrac{3times 5}{4times 4}cdotfrac{4times 6}{5times 5}cdots$
Do you see it?
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
add a comment |
$frac{1times 3}{2times 2}cdotfrac{2times 4}{3times 3}cdotfrac{3times 5}{4times 4}cdotfrac{4times 6}{5times 5}cdots$
Do you see it?
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
add a comment |
$frac{1times 3}{2times 2}cdotfrac{2times 4}{3times 3}cdotfrac{3times 5}{4times 4}cdotfrac{4times 6}{5times 5}cdots$
Do you see it?
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
$frac{1times 3}{2times 2}cdotfrac{2times 4}{3times 3}cdotfrac{3times 5}{4times 4}cdotfrac{4times 6}{5times 5}cdots$
Do you see it?
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
answered Nov 20 '18 at 16:55
TonyK
41.6k353132
41.6k353132
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
add a comment |
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
Nice answer tony the answer comes to 1/2
– Akash Roy
Nov 21 '18 at 17:42
add a comment |
hint
Take logarithm and use
$$
sum_{k=2}^{n+1}lnleft(1-frac{1}{k^2}right)
= sum_{k=2}^{n+1}left(lnleft(frac{k-1}{k}right)+lnleft(frac{k+1}{k}right)right)
=lnleft(frac{1}{n+1}right)+lnleft(frac{n+2}{2}right).
$$
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
add a comment |
hint
Take logarithm and use
$$
sum_{k=2}^{n+1}lnleft(1-frac{1}{k^2}right)
= sum_{k=2}^{n+1}left(lnleft(frac{k-1}{k}right)+lnleft(frac{k+1}{k}right)right)
=lnleft(frac{1}{n+1}right)+lnleft(frac{n+2}{2}right).
$$
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
add a comment |
hint
Take logarithm and use
$$
sum_{k=2}^{n+1}lnleft(1-frac{1}{k^2}right)
= sum_{k=2}^{n+1}left(lnleft(frac{k-1}{k}right)+lnleft(frac{k+1}{k}right)right)
=lnleft(frac{1}{n+1}right)+lnleft(frac{n+2}{2}right).
$$
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
hint
Take logarithm and use
$$
sum_{k=2}^{n+1}lnleft(1-frac{1}{k^2}right)
= sum_{k=2}^{n+1}left(lnleft(frac{k-1}{k}right)+lnleft(frac{k+1}{k}right)right)
=lnleft(frac{1}{n+1}right)+lnleft(frac{n+2}{2}right).
$$
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
edited Nov 20 '18 at 17:34
Viktor Glombik
582425
582425
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 16:55
hamam_Abdallah
37.9k21634
37.9k21634
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
add a comment |
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
It doesn't help. There is no need for logarithms.
– TonyK
Nov 20 '18 at 17:00
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
@TonyK I disagree. Logarithm transforms product into sum, and one may be more used to telescoping in sums rather than in products, so this may help indeed.
– lisyarus
Nov 20 '18 at 17:01
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
Nonsense. Just look at all the other answers -- no special functions are needed, just simple multiplication and division.
– TonyK
Nov 20 '18 at 18:38
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
@TonyK I'm not saying logarithm is needed here. I'm saying it may help understand for someone. Anyway, the usage of logarithms in this problem is just a matter of taste. If you think the answer is wrong, please, spend some time explaining why. As written, the answer looks completely valid.
– lisyarus
Nov 20 '18 at 19:13
add a comment |
Since
$$frac{sin(pi x)}{pi x}=prod_{ngeq 1}left(1-frac{x^2}{n^2}right) $$
we simply have
$$ prod_{ngeq 2}left(1-frac{1}{n^2}right)=lim_{xto 1}frac{sin(pi x)}{pi x(1-x^2)}=lim_{zto 0}frac{-sin(pi z)}{-pi z(z+1)(z+2)}=color{red}{frac{1}{2}}. $$
This technique is interesting since it allows a quick evaluation of $prod_{ngeq 1}left(1-frac{1}{2n^2}right)$ and similar products, which are not telescopic.
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
add a comment |
Since
$$frac{sin(pi x)}{pi x}=prod_{ngeq 1}left(1-frac{x^2}{n^2}right) $$
we simply have
$$ prod_{ngeq 2}left(1-frac{1}{n^2}right)=lim_{xto 1}frac{sin(pi x)}{pi x(1-x^2)}=lim_{zto 0}frac{-sin(pi z)}{-pi z(z+1)(z+2)}=color{red}{frac{1}{2}}. $$
This technique is interesting since it allows a quick evaluation of $prod_{ngeq 1}left(1-frac{1}{2n^2}right)$ and similar products, which are not telescopic.
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
add a comment |
Since
$$frac{sin(pi x)}{pi x}=prod_{ngeq 1}left(1-frac{x^2}{n^2}right) $$
we simply have
$$ prod_{ngeq 2}left(1-frac{1}{n^2}right)=lim_{xto 1}frac{sin(pi x)}{pi x(1-x^2)}=lim_{zto 0}frac{-sin(pi z)}{-pi z(z+1)(z+2)}=color{red}{frac{1}{2}}. $$
This technique is interesting since it allows a quick evaluation of $prod_{ngeq 1}left(1-frac{1}{2n^2}right)$ and similar products, which are not telescopic.
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
Since
$$frac{sin(pi x)}{pi x}=prod_{ngeq 1}left(1-frac{x^2}{n^2}right) $$
we simply have
$$ prod_{ngeq 2}left(1-frac{1}{n^2}right)=lim_{xto 1}frac{sin(pi x)}{pi x(1-x^2)}=lim_{zto 0}frac{-sin(pi z)}{-pi z(z+1)(z+2)}=color{red}{frac{1}{2}}. $$
This technique is interesting since it allows a quick evaluation of $prod_{ngeq 1}left(1-frac{1}{2n^2}right)$ and similar products, which are not telescopic.
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
answered Nov 20 '18 at 19:07
Jack D'Aurizio
287k33279656
287k33279656
add a comment |
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{prod_{k = 1}^{n}bracks{1 - {1 over pars{k + 1}^{2}}}} =
overbrace{{prod_{k = 1}^{n}k over prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {1 over n + 1}}}
overbrace{{prod_{k = 1}^{n}pars{k + 2} over
prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {n + 2 over 2}}} \[5mm] = &
{1 over 2},{1 + 2/n over 1 + 1/n}
,,,stackrel{mrm{as} n to infty}
{Large to},,,bbx{1 over 2}
end{align}
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{prod_{k = 1}^{n}bracks{1 - {1 over pars{k + 1}^{2}}}} =
overbrace{{prod_{k = 1}^{n}k over prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {1 over n + 1}}}
overbrace{{prod_{k = 1}^{n}pars{k + 2} over
prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {n + 2 over 2}}} \[5mm] = &
{1 over 2},{1 + 2/n over 1 + 1/n}
,,,stackrel{mrm{as} n to infty}
{Large to},,,bbx{1 over 2}
end{align}
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{prod_{k = 1}^{n}bracks{1 - {1 over pars{k + 1}^{2}}}} =
overbrace{{prod_{k = 1}^{n}k over prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {1 over n + 1}}}
overbrace{{prod_{k = 1}^{n}pars{k + 2} over
prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {n + 2 over 2}}} \[5mm] = &
{1 over 2},{1 + 2/n over 1 + 1/n}
,,,stackrel{mrm{as} n to infty}
{Large to},,,bbx{1 over 2}
end{align}
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{prod_{k = 1}^{n}bracks{1 - {1 over pars{k + 1}^{2}}}} =
overbrace{{prod_{k = 1}^{n}k over prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {1 over n + 1}}}
overbrace{{prod_{k = 1}^{n}pars{k + 2} over
prod_{k = 1}^{n}pars{k + 1}}}^{ds{= {n + 2 over 2}}} \[5mm] = &
{1 over 2},{1 + 2/n over 1 + 1/n}
,,,stackrel{mrm{as} n to infty}
{Large to},,,bbx{1 over 2}
end{align}
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
edited Nov 27 '18 at 1:57
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
answered Nov 22 '18 at 19:52
Felix Marin
67.1k7107141
67.1k7107141
add a comment |
add a comment |
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3
I'd try $(1-frac1{n^2})=(1-frac1 n)(1+frac1 n)$ then try just partial product $prod_{n=2}^m (1-frac 1n)$ then partial product $prod_{n=2}^m(1+frac 1n)$ then combine the partial products
– kingW3
Nov 20 '18 at 16:49
2
Your product omits $left(1-frac1{16}right)$; is this an error, or is it meant to be e.g. over only primes?
– Steven Stadnicki
Nov 20 '18 at 16:49
@StevenStadnicki You're correct. Fixed
– Stefan Ivanovski
Nov 20 '18 at 16:51